What Is The Expected Cost Of Repair Of A Car Selected At Random

Theexponential distribution is often concerned with the amount of time until some specific issue occurs. For example, the corporeality of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long altitude business telephone calls, and the amount of time, in months, a car battery lasts. It can exist shown, too, that the value of the change that you take in your pocket or bag approximately follows an exponential distribution.

Values for an exponential random variable occur in the post-obit style. There are fewer big values and more small values. For example, the amount of coin customers spend in ane trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.

The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.

Case

AllowX = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the boilerplate amount of time equal to four minutes.

X is a continuous random variable since time is measured. It is given that μ = four minutes. To exercise any calculations, you must know m, the decay parameter.

[latex]{grand}=\frac{1}{\mu}[/latex]. Therefore, [latex]{m}=\frac{1}{iv}={0.25}[/latex]

The standard deviation, σ, is the same as the mean. μ = σ

The distribution note is X ~ Exp(m). Therefore, X ~ Exp(0.25).

The probability density function is f(x) = me ^–mx . The number e = 2.71828182846… Information technology is a number that is used often in mathematics. Scientific calculators have the key "e^x ." If you enter one for x, the calculator will display the value due east.

The curve is:

f(x) = 0.25e ^–0.25x where x is at least zero and m = 0.25.

For example, f(5) = 0.25e ^{−(0.25)(five)} = 0.072. The postal clerk spends five minutes with the customers. The graph is as follows:

Notice the graph is a declining curve. When x = 0,

f(x) = 0.25east ^(−0.25)(0) = (0.25)(1) = 0.25 = one thousand. The maximum value on the y-axis is m.

Try It

The corporeality of fourth dimension spouses store for anniversary cards can exist modeled by an exponential distribution with the boilerplate amount of fourth dimension equal to eight minutes. Write the distribution, state the probability density role, and graph the distribution.

Solution:

X ~ Exp(0.125); f(x) = 0.125e^–0.125x ;

 Example

Using the information in example 1, find the probability that a clerk spends 4 to 5 minutes with a randomly selected client.

The curve is:

10 ~ Exp(0.125); f(x) = 0.125e^–0.125x

a) Find P(4 < x < 5).

Solution:

The cumulative distribution function (CDF) gives the area to the left.

P(10 < ten) = 1 – e^–mx

P(x < 5) = i – e(–0.25)(5) = 0.7135 and P(10 < 4) = 1 – e ^(–0.25)(4) = 0.6321

You can do these calculations easily on a calculator.

 The probability that a postal clerk spends four to five minutes with a randomly selected client is P(four < x < 5) = P(x < 5) – P(x < four) = 0.7135 − 0.6321 = 0.0814.

On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter eastward^(–0.25*4) – e^(–0.25*five).

b) Half of all customers are finished within how long? (Find the 50^thursday percentile)

Solution:

Find the l^thursday percentile.

P(x < g) = 0.50, k = ii.viii minutes (reckoner or computer)

Half of all customers are finished inside ii.8 minutes.

You lot tin also do the calculation as follows:

P(x < chiliad) = 0.50 and P(ten < k) = 1 –eastward ^–0.25k

Therefore, 0.fifty = 1 − due east ^{−0.25chiliad} and e ^−0.25k = one − 0.50 = 0.5

Take natural logs: ln(east ^–0.251000 ) = ln(0.50). And then, –0.25thousand = ln(0.50)

Solve for k:  [latex]{k}=\frac{ln0.50}{-0.25}={0.25}=2.8[/latex] minutes

c) Which is larger, the hateful or the median?

Solution:

From function b, the median or 50^th percentile is two.8 minutes. The theoretical mean is four minutes. The hateful is larger.

 Instance

The number of days ahead travelers purchase their airline tickets can exist modeled by an exponential distribution with the average corporeality of fourth dimension equal to 15 days. Find the probability that a traveler volition purchase a ticket fewer than x days in advance. How many days do half of all travelers wait?

Solution:

P(x < 10) = 0.4866

50^th percentile = x.40

Example

On the average, a certain computer part lasts ten years. The length of fourth dimension the computer role lasts is exponentially distributed.

 a) What is the probability that a calculator part lasts more than 7 years?

Solution:Allow x = the amount of time (in years) a computer part lasts.

[latex]\mu = {10}[/latex] so m = [latex]\frac{ane}{\mu} = \frac{i}{10}={0.ten}[/latex]
P(x > seven). Describe the graph.

P(x > 7) = 1 – P(x < vii).

Since P(10 < x) = 1 –e^–mx and then P(X > x) = ane –(1 –eastward^–mx ) = e^-mx

P(x > 7) = e ^(–0.1)(7) = 0.4966. The probability that a computer part lasts more than vii years is 0.4966.

On the dwelling house screen, enter e^(-.1*seven).

b) On the average, how long would 5 figurer parts last if they are used one after another?

Solution:

On the average, one computer part lasts ten years. Therefore, 5 computer parts, if they are used ane correct later the other would last, on the average, (5)(10) = 50 years.

c) Lxxx percentage of figurer parts last at well-nigh how long?

Solution:

Detect the 80^th percentile. Draw the graph. Let k = the eighty^th percentile.

Solve for k: [latex]{k}=\frac{ln(1-0.80)}{-0.1}={16.i}[/latex]

Lxxx percentage of the reckoner parts last at most xvi.1 years.

Solution:

Detect P(9 < ten < xi). Draw the graph.

d) What is the probability that a figurer office lasts between nine and 11 years?

Solution:

P(9 < ten < eleven) = P(10 < 11) – P(x < 9) = (ane – e ^(–0.1)(11)) – (1 – due east ^{(–0.one)(9)}) = 0.6671 – 0.5934 = 0.0737. The probability that a reckoner part lasts between nine and eleven years is 0.0737.

Example

Suppose that the length of a telephone call, in minutes, is an exponential random variable with disuse parameter = 1 12 . If some other person arrives at a public phone just earlier you, find the probability that yous volition take to look more than five minutes. Allow X = the length of a phone call, in minutes.

What is m, μ, and σ? The probability that you must wait more than five minutes is _______ .

Solution:

yard =  [latex]\frac{one}{12}[/latex]

[latex]\mu [/latex] = 12

[latex]\sigma [/latex] = 12

P(x > 5) = 0.6592

Case

The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of thirty customers per hour go far at a store and the time between arrivals is exponentially distributed.

1. On average, how many minutes elapse between two successive arrivals?
2. When the store starting time opens, how long on average does it take for three customers to go far?
3. After a customer arrives, find the probability that information technology takes less than i minute for the next customer to arrive.
4. Afterwards a customer arrives, find the probability that it takes more than five minutes for the next customer to make it.
5. Seventy percent of the customers arrive within how many minutes of the previous customer?
6. Is an exponential distribution reasonable for this situation?

Solutions:

1. Since we expect thirty customers to arrive per hour (threescore minutes), we expect on boilerplate one customer to get in every two minutes on average.
2. Since 1 customer arrives every ii minutes on average, it will have vi minutes on boilerplate for iii customers to arrive.
3. Let 10 = the fourth dimension between arrivals, in minutes. Past part a, μ = 2, then k = 1 two = 0.v.
   Therefore, X ∼ Exp(0.v).The cumulative distribution function is P(10 < ten) = ane – e(–0.vten) ^eastward .Therefore P(Ten < one) = i – east^(–0.v)(1) ≈ 0.3935

4. 

5. P(X > 5) = one – P(Ten < 5) = 1 – (1 – due east ^(–5)(0.5)) = e^–2.5 ≈ 0.0821.

6. 
7. Nosotros want to solve 0.70 = P(X < x) for x.
   Substituting in the cumulative distribution function gives 0.lxx = ane – e ^–0.5x , so that due east ^–0.5x = 0.thirty. Converting this to logarithmic grade gives –0.fivex = ln(0.thirty), or x = 50 n ( 0.30 ) – 0.v ≈ 2.41 minutes.Thus, seventy percent of customers make it within 2.41 minutes of the previous customer.
8. This model assumes that a single customer arrives at a fourth dimension, which may not exist reasonable since people might shop in groups, leading to several customers arriving at the same fourth dimension. Information technology also assumes that the menstruum of customers does not change throughout the twenty-four hours, which is non valid if some times of the mean solar day are busier than others.

Memorylessness of the Exponential Distribution

In instance i, recall that the amount of time betwixt customers is exponentially distributed with a mean of two minutes (X ~ Exp (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has at present elapsed, it would seem to exist more likely for a customer to make it within the side by side minute. With the exponential distribution, this is not the case–the additional fourth dimension spent waiting for the next customer does non depend on how much time has already elapsed since the last customer. This is referred to as the memoryless belongings. Specifically, the memoryless property says that

P (10 > r + t | X > r) = P (Ten > t) for all r ≥ 0 and t ≥ 0

For example, if five minutes has elapsed since the terminal client arrived, then the probability that more than than one infinitesimal will elapse earlier the next client arrives is computed by using r = v and t = 1 in the foregoing equation.

P(X > 5 + 1 | X > 5) = P(X > 1) = e ( – 0.v ) ( 1 ) ≈ 0.6065.

This is the aforementioned probability every bit that of waiting more than 1 minute for a client to arrive after the previous arrival.

The exponential distribution is often used to model the longevity of an electrical or mechanical device. In case 1, the lifetime of a sure computer part has the exponential distribution with a hateful of ten years (10 ~ Exp(0.ane)). The memoryless property says that knowledge of what has occurred in the by has no effect on time to come probabilities. In this case information technology means that an old part is non whatever more than likely to break down at any item time than a brand new part. In other words, the function stays as good every bit new until it suddenly breaks. For example, if the part has already lasted ten years, and then the probability that it lasts another seven years is P(Ten > 17|X > 10) =P(X > 7) = 0.4966.

Example

Refer to example i, where the time a postal clerk spends with his or her client has an exponential distribution with a hateful of four minutes. Suppose a client has spent four minutes with a postal clerk. What is the probability that he or she volition spend at least an boosted three minutes with the postal clerk?

The disuse parameter of Ten is m = 14 = 0.25, and so X ∼ Exp(0.25).
The cumulative distribution function is P(Ten < 10) = 1 – e–0.25x. We desire to find P(X > 7|X > 4). The memoryless property says that P(10 > 7|Ten > 4) = P (X > iii), and so nosotros merely demand to find the probability that a customer spends more than iii minutes with a postal clerk.
This is P(Ten > 3) = one – P (X < iii) = 1 – (1 – e–0.25⋅3) = e–0.75 ≈ 0.4724.

Relationship betwixt the Poisson and the Exponential Distribution

There is an interesting human relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between 2 successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the fourth dimension between events is not affected by the times between previous events. If these assumptions agree, and so the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. Retrieve that if X has the Poisson distribution with mean λ, then [latex]P(X=k)=\frac{{\lambda}^{k}{eastward}^{-\lambda}}{grand!}[/latex]. Conversely, if the number of events per unit time follows a Poisson distribution, then the corporeality of time between events follows the exponential distribution. (k! = g*(yard-one*)(chiliad–2)*(one thousand-3)…3*ii*1)

Case

At a police station in a large city, calls come in at an average charge per unit of iv calls per infinitesimal. Assume that the time that elapses from i telephone call to the next has the exponential distribution. Have note that we are concerned only with the rate at which calls come up in, and we are ignoring the fourth dimension spent on the phone. We must likewise assume that the times spent betwixt calls are independent. This means that a particularly long delay between 2 calls does not mean that there will be a shorter waiting period for the adjacent call. Nosotros may then deduce that the total number of calls received during a fourth dimension period has the Poisson distribution.

1. Find the boilerplate time between ii successive calls.
2. Find the probability that after a telephone call is received, the next call occurs in less than 10 seconds.
3. Find the probability that exactly v calls occur within a minute.
4. Find the probability that less than v calls occur within a infinitesimal.
5. Find the probability that more than 40 calls occur in an 8-minute period.

Solutions:

1. On average there are four calls occur per minute, so 15 seconds, or [latex]\frac{fifteen}{threescore} [/latex]= 0.25 minutes occur between successive calls on boilerplate.
2. Let T = fourth dimension elapsed betwixt calls. From office a, [latex]\mu = {0.25} [/latex], so grand = [latex]\frac{ane}{0.25} [/latex] = 4. Thus, T ~ Exp(4).  The cumulative distribution function is P(T < t) = 1 – e ^–4t . The probability that the adjacent phone call occurs in less than ten seconds (ten seconds = i/half-dozen minute) is P(T < [latex]\frac{ane}{6}[/latex]) = 1 –  [latex]{due east}^{-4\frac{1}{vi}} \approx{0.4866} [/latex]
3. Let Ten = the number of calls per minute. Equally previously stated, the number of calls per infinitesimal has a Poisson distribution, with a mean of four calls per infinitesimal. Therefore, 10 ∼ Poisson(4), then P(X = five) = [latex]\frac{{4}^{5}{e}^{-four}}{v!}\approx[/latex] 0.1563. (5! = (v)(four)(three)(ii)(1))
4. Keep in mind that Ten must exist a whole number, so P(X < 5) = P(X ≤ four).

   To compute this, nosotros could have P(Ten = 0) + P(X = i) + P(X = 2) + P(X = three) + P(X = four).

   Using technology, we run into that P(X ≤ 4) = 0.6288.

5. Let Y = the number of calls that occur during an viii minute catamenia.

   Since there is an average of four calls per infinitesimal, there is an average of (viii)(4) = 32 calls during each eight infinitesimal period.

   Hence, Y ∼ Poisson(32). Therefore, P(Y > xl) = one – P (Y ≤ 40) = 1 – 0.9294 = 0.0707.

Concept Review

If 10 has an exponential distribution with hateful [latex]\mu[/latex] and so the decay parameter is [latex]m =\frac{one}{\mu}[/latex], and we write Ten ∼ Exp(m) where x ≥ 0 and m > 0 . The probability density function of X is f(x) =me^-mx (or equivalently [latex]f(10)=\frac{1}{\mu}{e}^{\frac{-ten}{\mu}}[/latex].The cumulative distribution role of X is P(X≤ x) = one – e ^–mx .

The exponential distribution has the memoryless property, which says that future probabilities do not depend on whatsoever past information. Mathematically, information technology says that P(X > x + yard|X > x) = P(10 > chiliad).

If T represents the waiting time betwixt events, and if T ∼ Exp(λ), and so the number of events 10 per unit time follows the Poisson distribution with hateful λ. The probability density function of [latex]P\left(X=k\correct)=\frac{\lambda^{k}}{east^{-\lambda}}k![/latex]. This may be computed using a TI-83, 83+, 84, 84+ calculator with the command poissonpdf(λ, k). The cumulative distribution function P(Ten ≤ thou) may exist computed using the TI-83, 83+,84, 84+ calculator with the command poissoncdf(λ, k).

Formula Review

Exponential: X ~ Exp(thousand) where m = the decay parameter

• pdf: f(x) = 1000[latex]{e}^{-mx}[/latex] where 10 ≥ 0 and one thousand > 0
• cdf: P(10 ≤ x) = 1 –[latex]{east}^{-mx}[/latex]
• mean [latex]\mu = \frac{ane}{m}[/latex]
• standard deviation σ = µ
• percentile, chiliad: yard = [latex]\frac{ln(\text{AreaToTheLeftOfK})}{-m}[/latex]
• Additionally
  • P(10 > x) = e ^(–mx)
  • P(a < Ten < b) = e ^(–ma) – e ^(–mb)
• Memoryless Property: P(Ten > 10 + k|10 > x) = P (Ten > one thousand)
• Poisson probability:P ( Ten = g ) =[latex]\frac{{\lambda}^{g}{east}^{-\lambda}}{thousand!}[/latex] with hateful [latex]\lambda[/latex]
• thousand! = k*(k-one)*(chiliad-two)*(k-three)…3*two*1

References

Data from the United States Census Bureau.

Data from World Earthquakes, 2013. Available online at http://www.earth-earthquakes.com/ (accessed June 11, 2013).

"No-hitter." Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/pitcher/No-hitter (accessed June 11, 2013).

Zhou, Rick. "Exponential Distribution lecture slides." Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf‎ (accessed June 11, 2013).

Source: https://courses.lumenlearning.com/introstats1/chapter/the-exponential-distribution/

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